Which would be a much easier calculation. Is this correct in its assumption?

Once again, if this is soo apparently obvious sorry for posting, but if anyone could give me a feasible reason why this is not easier but at the same time using calculus, please do tell.

Mathstud

I don't see any reason it shouldn't work, except for a little problem in how you put in your substitution: You didn't take the square root. When you do this you get exactly the same expression you would get if you plugged $\displaystyle x = r~sin(\theta)$ into the integral.

-Dan

Edit: There is also a problem with your integration limits. The integral finds the area between the curve and the x-axis. Your limits would suggest that you are trying to find the area of a sector (wedge) of the circle. These will obviously not have the same area.

Which would be a much easier calculation. Is this correct in its assumption?

Once again, if this is soo apparently obvious sorry for posting, but if anyone could give me a feasible reason why this is not easier but at the same time using calculus, please do tell.

Mathstud

I don't think I understood what you are trying to say. Are you trying to determine n in terms of a and b... or what?

btw, there are some errors in your post; the integrand after the parametrization should be just cosine theta without the square; Also, you have $\displaystyle x = b = r cos \theta$or $\displaystyle \theta = arccos(b/r)$. Similarily for the lower limit. I don't see how you've got pi/n and 0 for your limits of the new integral.

Please elaborate..

PS: It's really annoying to keep typing [math*] and then [/math*] and then type those silly words again... can we suggest this forum to invent a new ... ( (huh??? ) , what do you call this?)

You may hate me for this, but have you consider using geometry for this ?

Supposing the case that a is negative and b is positive, then you know that the area of a sector is $\displaystyle 1/2 r^2 \theta$ so in this case $\displaystyle \theta = \pi - \arccos a/r - \arccos b/r$

Math tags

It's really annoying to keep typing [math*] and then [/math*] and then type those silly words again.

When you open a new post window you will see a toolbar just above the text box. On the second line start from the right and go over three tools. You will see a $\displaystyle \Sigma$. Highlight the Latex code you want inside the tags and click the $\displaystyle \Sigma$. This will insert both tags for you.

You may hate me for this, but have you consider using geometry for this ?

Supposing the case that a is negative and b is positive, then you know that the area of a sector is $\displaystyle 1/2 r^2 \theta$ so in this case $\displaystyle \theta = \pi - \arccos a/r - \arccos b/r$

Then add the areas of the triangles.

Bobak

Am I missing something here? The area represented by the integral looks something like the picture below. It is NOT a sector of the circle!

When you open a new post window you will see a toolbar just above the text box. On the second line start from the right and go over three tools. You will see a $\displaystyle \Sigma$. Highlight the Latex code you want inside the tags and click the $\displaystyle \Sigma$. This will insert both tags for you.

btw, there are some errors in your post; the integrand after the parametrization should be just cosine theta without the square;

What are you talking about?

Since the orientation of the parameterization is counterclockwise

$\displaystyle A=\int_a^{b}x(\theta)y'(\theta)d\theta$

So it would be $\displaystyle \cos^2(\theta)$

Also, you have or

Where did you get this b stuff from?

I don't see how you've got pi/n and 0 for your limits of the new integral.

As for the limits of integration, how is it not that? The zero should be obvious, the $\displaystyle \frac{\pi}{n}$ would come from the amout of $\displaystyle \theta$ needed to complete $\displaystyle \frac{1}{n}$th of the circle?

So...am I missing something? I appreciate that you care enough to point out what you believe are errors, but you should probably double check. But don't fret! I have been guilty of this many-a-time.

As for the limits of integration, how is it not that? The zero should be obvious, the $\displaystyle \frac{\pi}{n}$ would come from the amout of $\displaystyle \theta$ needed to complete $\displaystyle \frac{1}{n}$th of the circle?

So...am I missing something? I appreciate that you care enough to point out what you believe are errors, but you should probably double check. But don't fret! I have been guilty of this many-a-time.

Ah, I'm sorry; that post was nonsense...
My brain was fried at that time and I didn't understand what you meant in your post (English is not my native, btw);

Am I missing something here? The area represented by the integral looks something like the picture below. It is NOT a sector of the circle!

-Dan

True but you can spilt it up into a sector and two triangles, refer to my attachment. To get the area of a sector you need to know the angle suspending it. to figure out the angles between the triangles and the x-axis you use just inverse trig ( $\displaystyle \arccos{ \frac{a}{r}}$ etc... ) the add the area of the triangles.

I am fairly confident this problem can be solved using geometry alone.

Which would be a much easier calculation. Is this correct in its assumption?

Once again, if this is soo apparently obvious sorry for posting, but if anyone could give me a feasible reason why this is not easier but at the same time using calculus, please do tell.

Mathstud

I apologize if am getting annoying, but I'm afraid that I still do not completely understand what you are trying to use 1/n of the area of the semi-circle and stuff.

What you are using is basically polar coordinates integration given by

$\displaystyle
\int \int r dr d\theta
= \int 1/2 r^2 d\theta
$
where the limits of the inner integral is from 0 to r and the limits of the outer integral is from 0 to pi for your area of the semicircle.

In that case your variable is $\displaystyle \theta$ and (you claimed) the upper limit of the integral is $\displaystyle \pi /n$ in your post...

however, from the limits of the original integral you have x = a for lower and x = b for upper limit. Now, the problem is that you are actually integrating with respect to y if my reasoning on your derivation of

$\displaystyle r^2 \int cos^2 \theta d\theta
$

is correct; if the limits of the integral of $\displaystyle \sqrt{r^2 - y^2} dy$ is c to d, then we have

I apologize if am getting annoying, but I'm afraid that I still do not completely understand what you are trying to use 1/n of the area of the semi-circle and stuff.

What you are using is basically polar coordinates integration given by

$\displaystyle
\int \int r dr d\theta
= \int 1/2 r^2 d\theta
$
where the limits of the inner integral is from 0 to r and the limits of the outer integral is from 0 to pi for your area of the semicircle.

In that case your variable is $\displaystyle \theta$ and (you claimed) the upper limit of the integral is $\displaystyle \pi /n$ in your post...

however, from the limits of the original integral you have x = a for lower and x = b for upper limit. Now, the problem is that you are actually integrating with respect to y if my reasoning on your derivation of

$\displaystyle r^2 \int cos^2 \theta d\theta
$

is correct; if the limits of the integral of $\displaystyle \sqrt{r^2 - y^2} dy$ is c to d, then we have

Where $\displaystyle A_n$ denotes the area of $\displaystyle \frac{1}{n}$th of the circle.

I hope this is clear now.

I have never seen this, so if you do have a grievance please let me know.

And as was stated eariler, the obvious downfall is that $\displaystyle \frac{1}{n}$th of the circle is measured by Area so we need to know the area to find [/tex]n[/tex]. So unless a specific case such as a quarter or semi circle is encountered, or if someone stated a question in a way such as "Compute the area of one-___th ofa circle" it is pretty much useless.

Mathstud.

P.S. Don't ever worry about annoying someone, for most of the time it is the best way to learn.